Question: Let $f(x, y) = 2y^2x$. Suppose $\vec{a} = (3, -1)$ and $\vec{v} = \left( -2, -2 \right)$. Find the directional derivative of $f(x, y)$ at $\vec{a}$ in the direction of $\vec{v}$. Do not normalize the direction vector for your calculation. $\dfrac{\partial f}{\partial v} = $
Explanation: The directional derivative of a function $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $\left( \nabla f (\vec{a}) \right) \cdot \vec{v}$. Let's find the gradient of $f$. $\nabla f = \left( \dfrac{\partial f}{\partial x}, \dfrac{\partial f}{\partial y} \right) = \left( 2y^2, 4xy \right)$ Therefore: $\begin{aligned} \left( \nabla f (\vec{a}) \right) \cdot \vec{v} &= \left( 2, -12 \right) \cdot \left( -2, -2 \right) \\ \\ &= 20 \end{aligned}$ In conclusion, the directional derivative of $f$ at $\vec{a}$ in the direction of $\vec{v}$ equals $20$.